Balti Tee 2020 · Ülesanne 14

Let $M$ be the midpoint of $BC$. We first show that $X$ lies on $AM$. Consider $A'$, the reflection of $A$ across $M$. As $ABA'C$ is a parallelogram, we have that $\angle BA'C = \angle BAC = 180^\circ - \angle BHC$, which in turn gives us that $A'$ lies on $\omega$. Now $\angle HBA' = \angle HBC + \angle CBA' = \angle HBC + \angle ACB = 90^\circ$. Hence $HA$ is a diameter of $\omega$. In particular we must have $\angle HXA' = 90^\circ$. Consequently $\angle AXA' = \angle AXH + \angle HXA' = 90^\circ + 90^\circ = 180^\circ$, i.e. $A, X, A'$ are collinear. But $A, M, A'$ collinear by definition, hence $X$ lies on the $A$-median.

Now it suffices to show that $\angle AYZ = \angle AMZ$. We note the two following facts:

• $\angle AHX = \angle AYX$, since $\omega$ and $\Gamma$ have the same radius and the two angles span the same chord $AX$.
• $\omega$ is the reflection of the circumcircle of $ABC$ across $BC$. That gives us that $Z$ is the reflection of $A$ across $D$, the feet of the $A$-altitude to $BC$.

Hence we can write: $\angle AYZ = \angle AYX + \angle XYZ = \angle AHX + (180^\circ - \angle XHZ) = 2\angle AHX = 2\angle AMD = \angle AMZ$, which is what we wanted.